Soybean meal is 14% protein; cornmeal is 7% protein. How many pounds of each should be mixed together in order to get 280 lb mixture that is 9% protein?
How many pound of cornmeal should be in the mixture?
How many pounds of soybean meal should be in mixture?
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To solve this problem, we can use a method called the "mixture problem," where we determine the amount of each ingredient needed to achieve a desired mixture.
Let's start by defining the variables:
Let x represent the number of pounds of soybean meal needed.
Let y represent the number of pounds of cornmeal needed.
We have two pieces of information:
1. The total weight of the mixture is 280 pounds. So, we have the equation:
x + y = 280
2. The desired protein percentage in the mixture is 9%. We need to calculate the total amount of protein contributed by each ingredient and set it equal to 9% of the total weight.
The soybean meal is 14% protein, so the amount of protein from soybean meal is 0.14x pounds.
The cornmeal is 7% protein, so the amount of protein from cornmeal is 0.07y pounds.
The total amount of protein in the mixture is 0.09 * 280 pounds = 25.2 pounds.
This gives us the equation:
0.14x + 0.07y = 25.2
Now we have a system of two equations:
Equation 1: x + y = 280
Equation 2: 0.14x + 0.07y = 25.2
We can solve this system of equations to find the values of x and y.
One way to solve this is by using the substitution method:
Solve Equation 1 for x: x = 280 - y
Substitute this expression for x in Equation 2:
0.14(280 - y) + 0.07y = 25.2
Simplify and solve for y:
39.2 - 0.14y + 0.07y = 25.2
Combine like terms:
-0.07y = -14
Divide both sides by -0.07:
y = 200
So, you need 200 pounds of cornmeal in the mixture.
To find the amount of soybean meal, substitute the value of y back into Equation 1:
x + 200 = 280
x = 280 - 200
x = 80
Therefore, you need 80 pounds of soybean meal in the mixture.
Answer:
Pounds of cornmeal in the mixture: 200 pounds
Pounds of soybean meal in the mixture: 80 pounds